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QUESTION


Let $A$ be the $3\times3$ symmetric matrix

$$A=\left(\begin{array}{ccc}2&amp;1&amp;0\\1&amp;3&amp;1\\0&amp;1&amp;2\end{array}\right).$$

Calculate the determinant of the matrix $A$. Find the eigenvalues
of $A$, and construct the corresponding normalised eigenvectors.

\begin{description}

\item[(i)]
Show that your eigenvalues are mutually orthogonal.

\item[(ii)]
Write down an orthogonal matrix $R$ such that $R^TAR$ is diagonal
with the eigenvalues of $A$ as its diagonal entries. Verify this
by calculating $AR$ and $R^TAR$.

\end{description}




ANSWER


Determinant of $A$:

\begin{eqnarray*}
\left|\begin{array}{ccc}2&amp;1&amp;0\\1&amp;3&amp;1\\0&amp;1&amp;2\end{array}\right|&amp;=&amp;2\left|\begin{array}{cc}3&amp;1\\1&amp;2\end{array}\right|-\left|\begin{array}{cc}1&amp;1\\0&amp;2\end{array}\right|+0\\
&amp;=&amp;2(6-1)-(2)=8
\end{eqnarray*}

Eigenvalues correspond to solutions of

\begin{eqnarray*}
\left|\begin{array}{ccc}2-\lambda&amp;1&amp;0\\1&amp;3-\lambda&amp;1\\0&amp;1&amp;2-\lambda\end{array}\right|&amp;=&amp;0\\
(2-\lambda)\left[(3-\lambda)(2-\lambda)-1\right]-[2-\lambda]&amp;=&amp;0\\
(2-\lambda)(\lambda^2-5\lambda+2-2)&amp;=&amp;0\\
(2-\lambda)(\lambda^2-5\lambda+4)&amp;=&amp;0\\
(2-\lambda)(\lambda-1)(\lambda-4)&amp;=&amp;0
\end{eqnarray*}

Therefore the eigenvalues are 1, 2 and 4.

$\lambda=1$

$$\left(\begin{array}{ccc}1&amp;1&amp;0\\1&amp;2&amp;1\\0&amp;1&amp;1\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{c}0\\0\\0\end{array}\right)\ \
\left.\begin{array}{r}x+y=0\\x+2y+z=0\\y+z=0\end{array}\right\}
\begin{array}{c}x=-y\\z=-y\end{array}$$

A suitable eigenvector
$\mathbf{x}_1=\left(\begin{array}{c}1\\-1\\1\end{array}\right)$
which is normalised to
$\left(\begin{array}{c}\frac{1}{\sqrt{3}}\\-\frac{1}{\sqrt{3}}\\
\frac{1}{\sqrt{3}}\end{array}\right)$.


$\lambda=2$

$$\left(\begin{array}{ccc}0&amp;1&amp;0\\1&amp;1&amp;1\\0&amp;1&amp;0\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{c}0\\0\\0\end{array}\right)\ \
\left.\begin{array}{r}y=0\\x+y+z=0\\y=0\end{array}\right\}
\begin{array}{c}y=0\\z=-x\end{array}$$

A suitable eigenvector
$\mathbf{x}_2=\left(\begin{array}{c}1\\0\\-1\end{array}\right)$
which is normalised to
$\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\0\\
-\frac{1}{\sqrt{2}}\end{array}\right)$.


$\lambda=4$

$$\left(\begin{array}{ccc}-2&amp;1&amp;0\\1&amp;-1&amp;1\\0&amp;1&amp;-2\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{c}0\\0\\0\end{array}\right)\ \
\left.\begin{array}{r}-2x+y=0\\x-y+z=0\\y-2z=0\end{array}\right\}
\begin{array}{c}y=2x\\y=2z\end{array}$$

A suitable eigenvector
$\mathbf{x}_1=\left(\begin{array}{c}1\\2\\1\end{array}\right)$
which is m=normalised to
$\left(\begin{array}{c}\frac{1}{\sqrt{6}}\\\frac{2}{\sqrt{6}}\\
\frac{1}{\sqrt{6}}\end{array}\right)$.


\begin{description}

\item[(i)]
Mutually orthogonal eigenvectors $\mathbf{x}_i,\mathbf{x}_j$
satisfy $\mathbf{x}_i.\mathbf{x}_j=0=\mathbf{x}_j\mathbf{x}_i$

$\left.\begin{array}{c}\mathbf{x}_1.\mathbf{x}_2=
\left(\begin{array}{c}1\\-1\\1\end{array}\right)
\left(\begin{array}{c}1\\0\\-1\end{array}\right) =1+0-1=0\\
\mathbf{x}_1.\mathbf{x}_3=
\left(\begin{array}{c}1\\-1\\1\end{array}\right)
.\left(\begin{array}{c}1\\2\\1\end{array}\right) =1-2+1=0\\
\mathbf{x}_2.\mathbf{x}_3=
\left(\begin{array}{c}1\\0\\-1\end{array}\right)
.\left(\begin{array}{c}1\\2\\1\end{array}\right)=1+0-1=0
\end{array}\right\}$ so eigenvectors are mutually orthogonal.

\item[(ii)]

$$R=\left(\begin{array}{ccc}\frac{1}{\sqrt{3}}&amp;\frac{1}{\sqrt{2}}&amp;\frac{1}{\sqrt{6}}\\
-\frac{1}{\sqrt{3}}&amp;0&amp;\frac{2}{\sqrt{6}}\\
\frac{1}{\sqrt{3}}&amp;-\frac{1}{\sqrt{2}}&amp;\frac{1}{\sqrt{6}}\end{array}\right)\
\
R^T=\left(\begin{array}{ccc}\frac{1}{\sqrt{3}}&amp;-\frac{1}{\sqrt{3}}&amp;\frac{1}{\sqrt{3}}\\
\frac{1}{\sqrt{2}}&amp;0&amp;-\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{6}}&amp;\frac{2}{\sqrt{6}}&amp;\frac{1}{\sqrt{6}}\end{array}\right)$$

\begin{eqnarray*}
R^TAR&amp;=&amp;\left(\begin{array}{ccc}1&amp;0&amp;0\\0&amp;2&amp;0\\0&amp;0&amp;4\end{array}\right)\\
AR&amp;=&amp;\left(\begin{array}{ccc}2&amp;1&amp;0\\1&amp;3&amp;1\\0&amp;1&amp;2\end{array}\right)
\left(\begin{array}{ccc}\frac{1}{\sqrt{3}}&amp;\frac{1}{\sqrt{2}}&amp;\frac{1}{\sqrt{6}}\\
-\frac{1}{\sqrt{3}}&amp;0&amp;\frac{2}{\sqrt{6}}\\
\frac{1}{\sqrt{3}}&amp;-\frac{1}{\sqrt{2}}&amp;\frac{1}{\sqrt{6}}\end{array}\right)\\
&amp;=&amp;\left(\begin{array}{ccc}\frac{1}{\sqrt{3}}&amp;\frac{2}{\sqrt{2}}&amp;\frac{4}{\sqrt{6}}\\
-\frac{1}{\sqrt{3}}&amp;0&amp;\frac{8}{\sqrt{6}}\\
\frac{1}{\sqrt{3}}&amp;-\frac{2}{\sqrt{2}}&amp;\frac{4}{\sqrt{6}}\end{array}\right)\\
R^TAR&amp;=&amp;\left(\begin{array}{ccc}\frac{1}{\sqrt{3}}&amp;-\frac{1}{\sqrt{3}}&amp;\frac{1}{\sqrt{3}}\\
\frac{1}{\sqrt{2}}&amp;0&amp;-\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{6}}&amp;\frac{2}{\sqrt{6}}&amp;\frac{1}{\sqrt{6}}\end{array}\right)
\left(\begin{array}{ccc}\frac{1}{\sqrt{3}}&amp;\frac{2}{\sqrt{2}}&amp;\frac{4}{\sqrt{6}}\\
-\frac{1}{\sqrt{3}}&amp;0&amp;\frac{8}{\sqrt{6}}\\
\frac{1}{\sqrt{3}}&amp;-\frac{2}{\sqrt{2}}&amp;\frac{4}{\sqrt{6}}\end{array}\right)
=\left(\begin{array}{ccc}1&amp;0&amp;0\\0&amp;2&amp;0\\0&amp;0&amp;4\end{array}\right)
\end{eqnarray*}

\end{description}




\end{document}
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